Why is lightning so short?

The diagram shows the trajectory of a ball that

11Alexandr11 [23.1K]

Answer:

69 MPH

Explanation:

4 0

3 years ago

A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured

nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0

4 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

3 years ago

If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.

horrorfan [7]

This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4

8 0

4 years ago

Read 2 more answers